What are three steps scientist take to evaluate a scientific explanation

Please help me. Can you explain energy principle level?

Likurg_2 [28]

<span>Rules for the Principal Energy Level. A principal energy level may contain up to 2n2 electrons, with n being the number of each level. The set of three p orbitals can hold up to 6 electrons. Thus, the second principal energy level can hold up to 8 electrons, 2 in the s orbital and 6 in the p orbital.</span>

8 0

3 years ago

How many molecules are in 1 mole of nitrogen sulfide

gulaghasi [49]

Answer:

46.07

Explanation:

im not 100 sure this is right almost tho

please mark brainliest

6 0

3 years ago

A solution is prepared by condensing 4.00 L of a gas,

vredina [299]

Answer:

-2.3 ºC

Explanation:

Kf (benzene) = 5.12 ° C kg mol – 1

1st - We calculate the moles of condensed gas using the ideal gas equation:

n = PV / (RT)

P = 748/760 = 0.984 atm

T = 270 + 273.15 = 543.15 K

V = 4 L

R = 0.082 atm.L / mol.K

n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol

Then, you calculate the molality of the solution:

m = n / kg solvent

m = 0.088 mol / 0.058 kg = 1.52mol / kg

Then you calculate the decrease in freezing point (DT)

DT = m * Kf

DT = 1.52 * 5.12 = 7.8 ° C

Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:

T = 5.5 - 7.8 = -2.3 ºC

5 0

3 years ago

What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed

monitta

Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH = 1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

= 32.042 g/mol

Molar Mass of CO₂ = 1(12.01 g/mol) + 2(16.00 g/mol)

= 44.01 g/mol

Mass of only one reactant i.e. CH₃OH is given so it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

2 mol CO₂ / 2 mol CH₃OH

Convert moles of CO₂ to grams of CO₂ utilizing molar mass of CO₂ as:

44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ] [1 mol CH₃OH / 32.042 g CH₃OH ] [2 mol CO₂ / 2 mol CH₃OH] [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

= 14.28 g CO₂

Theoretical yield = 14.28 g CO₂

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

= 0.707 x 100%

= 70.7 %

Hence option A 70.7% yield is the correct answer.

8 0

3 years ago

Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.

OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = $\frac{69} {100} x 1420 g$

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = $\frac{979.8 g}{63 g/ mol}$

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = $\frac{15.55 x 1000}{440.2}$

= 35.325 molal

3 0

3 years ago